Let $y=\dfrac{\cos(x)}{x^3}$. $\dfrac{dy}{dx}=$
Answer: $\dfrac{\cos(x)}{x^3}$ is the quotient of two, more basic, expressions: $\cos(x)$ and $x^3$. Therefore, $\dfrac{dy}{dx}$ can be found using the quotient rule : $\begin{aligned} \dfrac{d}{dx}\left[\dfrac{u(x)}{v(x)}\right]&=\dfrac{\dfrac{d}{dx}[u(x)]v(x)-u(x)\dfrac{d}{dx}[v(x)]}{[v(x)]^2} \\\\ &=\dfrac{u'(x)v(x)-u(x)v'(x)}{[v(x)]^2} \end{aligned}$ Let's differentiate! $\begin{aligned} &\phantom{=}\dfrac{dy}{dx} \\\\ &=\dfrac{d}{dx}\left(\dfrac{\cos(x)}{x^3}\right) \\\\ &=\dfrac{\dfrac{d}{dx}(\cos(x))x^3-\cos(x)\dfrac{d}{dx}(x^3)}{(x^3)^2}&&\gray{\text{The quotient rule}} \\\\ &=\dfrac{-\sin(x)\cdot x^3-\cos(x)\cdot 3x^2}{(x^3)^2}&&\gray{\text{Differentiate }\cos(x)\text{ and }x^3} \\\\ &=-\dfrac{x^2(x\sin(x)+3\cos(x))}{x^6}&&\gray{\text{Simplify}} \\\\ &=-\dfrac{x\sin(x)+3\cos(x)}{x^4}&&\gray{\text{Cancel common factors}} \end{aligned}$ In conclusion, $\dfrac{dy}{dx}=-\dfrac{x\sin(x)+3\cos(x)}{x^4}$ or any other equivalent form.